in $\Bbb Q[x]$; if we denote the ideal $\langle X^4 - 5X^2 +6 \rangle \subset \Bbb Q[x]$ by $J$, this implies that

$((X^2 - 2) + J)((X^2 - 3) + J) = 0 \tag{2}$

in $\Bbb Q[x] / J$, since

$(X^2 - 2)(X^2 - 3) = X^4- 5X^2 + 6 \in J. \tag{3}$

But

$(X^2 - 2) + J \ne 0 \ne (X^2 - 3) + J \tag{4}$

in $\Bbb Q[x]/J$, since

$\deg X^2 - 2 = \deg X^2 - 3 = 2, \tag{5}$

prohibiting

$X^2 - 2, X^2 - 3 \in J, \tag{6}$

for neither can then be a multiple of $X^4 - 5X^2 + 6$ in $\Bbb Q[x]$. Thus both $X^2 - 2$ and $X^2 - 3$ are zero divisors in $\Bbb Q[x] / J$

In general, on can seek out zero divisors in $\Bbb F[x]/\langle p(x) \rangle$ where $p(x) \in \Bbb F[x]$ for any field $\Bbb F$ by looking for non-trivial factors of $p(x)$ in $\Bbb F[x]$. If $p(x) = q_1(x)q_2(x) \in \Bbb F[x]$, where $\deg q_1, \deg q_2 \ge 1$, then